Question

A baseball is popped straight up into the air and has a hang-time of 6 s. Determine the maximum height to which the ball rises?

• A. 180 m
• B. 90 m
• C. 45 m
• D. 22.5 m

Answer 1

The correct option is C

45 m

Time taken to reach the maximum height = $\frac{6}{2}=3s$
Final velocity, $v=0$, $a=-10m{s}^{-2}$( since moving upwards )
Using the first equation of motion, $v=u+at$,we get,
$0=u-10\times 3$
$\Rightarrow$$u=30m{s}^{-2}$
From second equation of motion, we have ,s = ut + $\frac{1}{2}a{t}^2$
$\Rightarrow$ $s=30\times 3$ – $\frac{1}{2}\times 10\times 3\times 3$
$\Rightarrow$ $s=90–45=45m$